Maths Help (1 Viewer)

Farhanthestudent005 · Dec 4, 2021

Will appreciate it if I am helped with this, thanks....

jimmysmith560 · Dec 4, 2021

Would this help?

Part (a):

Part (b):

CM_Tutor · Dec 5, 2021

Though not the method requested, it is also possible to shift the axis of rotation $y = 10$ down by 10 units, so that it becomes the $x$ -axis. The shifted parabola is then $y = 2x^2 - 10$ . Then, using usual volumes by revolution:

$\begin{align*} V_{\text{parabola}} = \pi\int y^2\ dx &= \pi\int_{-2}^2 \left(2x^2 - 10\right)^2\ dx \\ &= 4\pi\int_{-2}^2 x^4 - 10x^2 + 25\ dx \\ &= 8\pi\int_0^2 x^4 - 10x^2 + 25\ dx \qquad \text{as the integrand is an EVEN function} \\ &= 8\pi\left[\cfrac{x^5}{5} - \cfrac{10x^3}{3} + 25x\right]_0^2 \\ &= 8\pi\left[\left(\cfrac{2^5}{5} - \cfrac{10(2)^3}{3} + 25(2)\right) - \left(\cfrac{0^5}{5} - \cfrac{10(0)^3}{3} + 25(0)\right)\right] \\ &= 8\pi\left[\left(\cfrac{32 \times 3}{15} - \cfrac{80\times 5}{15} + 50\right) - 0\right] \\ &= 8\pi \times \cfrac{446}{15} \\ &= \cfrac{3568\pi}{15}\ \text{cubic units} \end{align*}$

But, the volume found here includes the region between the parabolic shape and the axis of rotation (which is a cylinder) and which must be removed to find the volume of the required solid:

$V_{\text{cylinder}} = \pi r^2 h = \pi \times 2^2 \times 4 = 16\pi\ \text{cubic units}$

$\text{So,} \quad V_{\text{solid}} = V_{\text{parabola}} - V_{\text{cylinder}} = \cfrac{3568\pi}{15} - 16\pi = \cfrac{3328\pi}{15}\ \text{cubic units}$

Farhanthestudent005 · Dec 10, 2021

Yes thank you it helped, I greatly appreciate it

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Maths Help (1 Viewer)

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